Question

The half-life of Tungsten-181 is 121 days. If you start with 3 pounds of it, how much do you have after 7 years?

Answer 1

Hint: For solving this question, we need to use the concept of exponential decay. We can say that a quantity is subject to exponential decay if it decreases at a rate proportional to its current value. Therefore, here we will use the decay formula for determining our answer.
Formula used:
 \[M = {M_0}{\left( {\dfrac{1}{2}} \right)^n}\] \[\] , where, \[M\] is the final mass, \[{M_0}\] is the initial mass and \[n\] is the number of half-lives.

Complete step by step solution:
Here, we are given the duration of 7 years.
Let us first convert it into days as the half-life of Tungsten-181 is given in the days.
We know that $ 7year's = 365 \times 7 = 2555days $ .
We are given that the half-life of Tungsten-181 is 121 days.
Now, we will find the half-lives of Tungsten-181 passed during the period of 7 years.
 $ \Rightarrow n = \dfrac{{2555}}{{121}} = 21.12 $
Now we will use the equation \[M = {M_0}{\left( {\dfrac{1}{2}} \right)^n}\] .
The initial mass is given as 3 pounds. Therefore, $ {M_0} = 3pounds $ .
We have also determined the number of half-lives during the given period which is $ n = 21.12 $ .
 \[M = {M_0}{\left( {\dfrac{1}{2}} \right)^n} = 3 \times {\left( {\dfrac{1}{2}} \right)^{21.12}} = 1.33 \times {10^{ - 6}}pounds\] .
Thus, after 7 years, we will have \[1.33 \times {10^{ - 6}}pounds\] of Tungsten-181
So, the correct answer is “ \[1.33 \times {10^{ - 6}}pounds\] ”.

Note: Here, we have used the concept of exponential decay. Exponential decay is different from linear decay because decay factor relies on a percentage of the original amount, which means the actual number the original amount might be reduced by will change over time whereas a linear function decreases the original number by the same amount every time.